AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.1

### 10th Class Maths 1st Lesson Real Numbers Ex 1.1 Textbook Questions and Answers

Question 1.
Use Euclid’s division algorithm to find the HCF of
i) 900 and 270
900 = 270 × 3 + 90
270 = 90 × 3 + 0
∴ HCF = 90

ii) 196 and 38220
38220 = 196 × 195 + 0
∴ 196 is the HCF of 196 and 38220.

iii) 1651 and 2032
2032 = 1651 × 1 + 381
1651 = 381 × 4 + 127
381 = 127 × 3 + 0
∴ HCF = 127

Question 2.
Use Euclid division lemma to show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integers.
Let ‘a’ be an odd positive integer.
Let us now apply division algorithm with a and b = 6.
∵ 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5.
i.e., ’a’ can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient.
But ‘a’ is taken as an odd number.
∴ a can’t be 6q or 6q + 2 or 6q + 4.
∴ Any odd integer is of the form 6q + 1, 6q + 3 or 6q + 5.

Question 3.
Use Euclid’s division lemma to show that the square of any positive integer is of the form 3p, 3p + 1.
Let ‘a’ be the square of an integer.
Applying Euclid’s division lemma with a and b = 3
Since 0 ≤ r < 3, the possible remainders are 0, 1, and 2.
∴ a = 3q (or) 3q + 1 (or) 3q + 2
∴ Any square number is of the form 3q, 3q + 1 or 3q + 2, where q is the quotient.
(or)
Let ‘a’ be a positive integer
So it can be expressed as a = bq + r (from Euclideans lemma)
now consider b = 3 then possible values of ‘r’ are ‘0’ or ‘1’ or 2.
then a = 3q + 0 = 3q (or) 3q + 1 or 3q + 2 now square of given positive integer (a2) will be
Case – I: a2 – (3q)2 = 9q2=3(3q2) = 3p (p = 3q2)
Case-II: a2 = (3q + l)2 = 9q2 + 6q+ 1
= 3[3q2 + 2q] + 1 = 3p+l (Where p = 3q2 + 2q) or
Case – III: a2 = (3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1
= 3[3q2 + 4q + 1] + 1
= 3p + 1 (where ‘p’ = 3q2 + 4q + 1)
So from above cases 1, 2, 3 it is clear that square of a positive integer (a) is of the form 3p or 3p + 1
Hence proved.

Question 4.
Use Euclid’s division lemma to show that the cube of a positive integer is of the form 9m, 9m + 1 or 9m + 8.
(OR)
Show that the cube of any positive integer is of form 9m or 9m + 1 or 9m + 8, where m is an integer.
Let ‘a’ be positive integer. Then from Euclidean lemma a = bq + r;
now consider b = 9 then 0 ≤ r < 9, it means remainder will be 0, or 1, 2, 3, 4, 5, 6, 7, or 8
So a = bq + r
⇒ a = 9q + r (for b = 9)
now cube of a = a3 + (9q + r)3
= (9q)3 + 3.(9q)3r + 3. 9q.r + r3
= 93q3 + 3.92(q2r) + 3.9(q.r) + r3
= 9[92.q3 + 3.9.q2r + 3.q.r] + r3
a3 = 9m + r3 (where ‘m’ = 92q3 + 3.9.q2r + 3.q.r)
if r = 0 ⇒ r3 = 0 then a3 = 9m + 0 = 9m
and for r = 1 ⇒ r3 = l3 then a3 = 9m + 1
and for r = 2 ⇒ r3 = 23 then a3 = 9m + 8
for r = 3 ⇒ r3, = 33 ⇒ a3 = 9m + 27 = 9(m) where m = (9m +3)
for r = 4 ⇒ r3 = 43 ⇒ a3 = 9m + 64 = (9m + 63) + 1 = 9m + 1
for r = 5 ⇒ r3 = 125 ⇒ a3 = 9m + 125 = (9m + 117) + 8 = 9m + 8
for r = 6 ⇒ r3 — 216 ⇒ a3 = 9m + 216 = 9m + 9(24) = 9m
for r = 7 ⇒ r3 = 243
⇒ a3 = 9m + 9(27) = 9m
for r = 8 ⇒ r3 = 512
⇒ a3 = 9m + 9(56) + 8 = 9m + 8
So from the above it is clear that a3 is either in the form of 9m or 9m + 1 or 9m + 8.
Hence proved.

Question 5.
Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.
(Or)
Show that one and only one out of a, a + 2 and a + 4 is divisible by 3 where ‘a’ is any positive integer.
Let ‘n’ be any positive integer.
Then from Euclidean’s lemma n = bq + r (now consider b = 3)
⇒ n = 3q + r (here 0 ≤ r < 3) which means the possible values of ‘r’ = 0 or 1 or 2
Now consider r = 0 then ‘n’ = 3q (divisible by 3)
and n + 2 = 3q + 2 (not divisible by 3)
n + 4 = 3q + 4 (not divisible by 3)
Case – II: For r = 1
n = 3q + 1 (not divisible by 3)
n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + l) divisible by 3
n + 4 = 3q + 1 + 4 = 3q + 5 not divisible by 3
Case – III: For r = 2,
n = 3q + 2 not divisible by 3
n + 2 = 3q + 2 + 2 = 3q + 4, not divisible by 3
n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) divisible by 3
So in all above three cases we observe, only one of either (n) or (n + 1) or (n + 4) is divisible by 3.
Hence proved.